Question

If $k=\tan {{25}^{\circ }}$ then find $\dfrac{k-1}{k+1}+\dfrac{k+1}{k-1}$.
\[\begin{align}
  & A.2\cos ec{{130}^{\circ }} \\
 & B.2\cos ec{{45}^{\circ }} \\
 & C.2\sec {{130}^{\circ }} \\
 & D.2\sec {{400}^{\circ }} \\
\end{align}\]

Answer 1

Hint: In this question, we are given k to be equal to $k=\tan {{25}^{\circ }}$ and we have to evaluate $\dfrac{k-1}{k+1}+\dfrac{k+1}{k-1}$. For this, we will first simplify the given expression in terms of k and then put values of k as $\tan {{25}^{\circ }}$. After this, we will use various trigonometric identities to simplify our answer and obtain the required result in terms of secant or cosecant. Various trigonometric identities that we will use are:
\[\begin{align}
  & \left( i \right)1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \\
 & \left( ii \right)\tan \theta =\dfrac{\sin \theta }{\cos \theta } \\
 & \left( iii \right)\cos \theta =\dfrac{1}{\sec \theta } \\
 & \left( iv \right){{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\cos 2\theta \\
 & \left( v \right)\cos \left( {{180}^{\circ }}-\theta \right)=-\cos \theta \\
\end{align}\]

Complete step by step answer:
Here we are given k as $\tan {{25}^{\circ }}$. And we need to find the value of $\dfrac{k-1}{k+1}+\dfrac{k+1}{k-1}$.
Let us first simplify the given expression before putting in the value of k.
Given expression is $\dfrac{k-1}{k+1}+\dfrac{k+1}{k-1}$.
Taking LCM (k+1)(k-1), we get:
\[\begin{align}
  & \Rightarrow \dfrac{\left( k-1 \right)\left( k-1 \right)+\left( k+1 \right)\left( k+1 \right)}{\left( k+1 \right)\left( k-1 \right)} \\
 & \Rightarrow \dfrac{{{\left( k-1 \right)}^{2}}+{{\left( k+1 \right)}^{2}}}{\left( k+1 \right)\left( k-1 \right)} \\
\end{align}\]
As we know, basic algebraic identities are ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab,{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\text{ and }\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$. So using them in above expression, we get:
\[\begin{align}
  & \Rightarrow \dfrac{{{k}^{2}}+1-2k+{{k}^{2}}+1+2k}{{{k}^{2}}-1} \\
 & \Rightarrow \dfrac{2{{k}^{2}}+2}{{{k}^{2}}-1} \\
 & \Rightarrow \dfrac{2\left( {{k}^{2}}+1 \right)}{{{k}^{2}}-1} \\
\end{align}\]
Now let us put the value of k as $\tan {{25}^{\circ }}$ we get:
\[\Rightarrow \dfrac{2\left( {{\tan }^{2}}{{25}^{\circ }}+1 \right)}{{{\tan }^{2}}{{25}^{\circ }}-1}\]
We know that, $1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta $ therefore, numerator of this expression will become $2{{\sec }^{2}}{{25}^{\circ }}$. Hence, we get:
\[\Rightarrow \dfrac{2{{\sec }^{2}}{{25}^{\circ }}}{{{\tan }^{2}}{{25}^{\circ }}-1}\]
Using $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ in denominator we get:
\[\Rightarrow \dfrac{2{{\sec }^{2}}{{25}^{\circ }}}{\dfrac{si{{n}^{2}}{{25}^{\circ }}}{{{\cos }^{2}}{{25}^{\circ }}}-1}\]
Taking LCM in denominator we get:
\[\Rightarrow \dfrac{2{{\sec }^{2}}{{25}^{\circ }}}{\dfrac{si{{n}^{2}}{{25}^{\circ }}-{{\cos }^{2}}{{25}^{\circ }}}{{{\cos }^{2}}{{25}^{\circ }}}}\]
Simplifying we get:
\[\Rightarrow \dfrac{2{{\sec }^{2}}{{25}^{\circ }}\cdot {{\cos }^{2}}{{25}^{\circ }}}{si{{n}^{2}}{{25}^{\circ }}-{{\cos }^{2}}{{25}^{\circ }}}\]
Now, we know $\cos \theta =\dfrac{1}{\sec \theta }$ so ${{\cos }^{2}}{{25}^{\circ }}=\dfrac{1}{{{\sec }^{2}}{{25}^{\circ }}}$ hence we get:
\[\begin{align}
  & \Rightarrow \dfrac{\dfrac{2{{\sec }^{2}}{{25}^{\circ }}}{{{\sec }^{2}}{{25}^{\circ }}}}{si{{n}^{2}}{{25}^{\circ }}-{{\cos }^{2}}{{25}^{\circ }}} \\
 & \Rightarrow \dfrac{2}{si{{n}^{2}}{{25}^{\circ }}-{{\cos }^{2}}{{25}^{\circ }}} \\
\end{align}\]
Taking negative sign common in denominator we get:
\[\Rightarrow \dfrac{2}{-\left( {{\cos }^{2}}{{25}^{\circ }}-si{{n}^{2}}{{25}^{\circ }} \right)}\]
Using ${{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\cos 2\theta $ in the denominator we get:
\[\Rightarrow \dfrac{2}{-\cos {{50}^{\circ }}}\]
Now we can see none of the options have negative signs, so changing the angle of $\cos \theta $ to remove negative signs. As we know $\cos \left( {{180}^{\circ }}-\theta \right)=-\cos \theta $ so putting $\theta ={{50}^{\circ }}$ we get \[-\cos {{50}^{\circ }}=\cos \left( {{180}^{\circ }}-\theta \right)=\cos {{130}^{\circ }}\], so expression becomes $\dfrac{2}{\cos {{130}^{\circ }}}$.
Using $\cos \theta =\dfrac{1}{\sec \theta }$ we get:
\[\begin{align}
  & \Rightarrow \dfrac{2}{\dfrac{1}{\sec {{130}^{\circ }}}} \\
 & \Rightarrow 2\sec {{130}^{\circ }} \\
\end{align}\]
Hence, for $k=\tan {{25}^{\circ }}$,$\dfrac{k-1}{k+1}+\dfrac{k+1}{k-1}$ will be equal to $2\sec {{130}^{\circ }}$.

So, the correct answer is “Option C”.

Note: Students should take care of signs while applying algebraic identities. Students should not get confused with the reciprocal of $\sin \theta \text{ and }\cos \theta $. Reciprocal of $\sin \theta $ is $\text{cosec}\theta $ and reciprocal of $\cos \theta $ is $\sec \theta $. Also take care of signs while using trigonometric identities such as $1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta ,{{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\cos 2\theta $. Students should remember properties such as $\cos \left( {{180}^{\circ }}-\theta \right),\cos \left( {{180}^{\circ }}+\theta \right),\sin \left( {{90}^{\circ }}+\theta \right)$ and others.